Answer
$\displaystyle \frac{dy}{dx}=0$
Work Step by Step
(At the end of secion 10-6, we found that $\displaystyle \frac{d}{dx}[\ |x|\ ]=\frac{|x|}{x}$ )
$f(x)=|x|,\displaystyle \ \ \quad g(x)=x,\quad y=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=\displaystyle \frac{|x|}{x}\qquad g^{\prime}(x)=1$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{f(x)}{g(x)}]$= ... quotient rule ...
$= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$=\displaystyle \frac{\frac{|x|}{x}\cdot x - |x|\cdot 1}{x^{2}}$
$=\displaystyle \frac{|x|-|x|}{x^{2}} =0$
